use quarto, not Pluto to render pages

CwJ/integral_vector_calculus/stokes_theorem.jmd

@@ -114,24 +114,24 @@ Let's consider now what an integral over the boundary would mean. The region, or
 
 With this choice of integral over the boundary, we can see much cancellation arises were we to compute this integral for each piece, as we would have with $a=x_0 < x_1 < \cdots x_{n-1} < x_n=b$:
 
-$$~
+```math
 (F(x_1) - F(x_0)) + (F(x_2)-F(x_1)) + \cdots + (F(x_n) - F(x_{n-1})) = F(x_n) - F(x_0) = F(b) - F(a).
-~$$
+```
 
 That is, with this definition for a boundary integral, the interior pieces of the microscopic approximation cancel and the total is just the integral over the oriented macroscopic boundary $\{a, b\}$.
 
 But each microscopic piece can be reimagined, as
 
-$$~
+```math
 F(x_{i}) - F(x_{i-1}) = \left(\frac{F(x_{i}) - F(x_{i-1})}{\Delta{x}}\right)\Delta{x}
 \approx F'(x_i)\Delta{x}.
-~$$
+```
 
 The approximation could be exact were the mean value theorem used to identify a point in the interval, but we don't pursue that, as the key point is the right hand side is a Riemann sum approximation for a *different* integral, in this case the integral $\int_a^b F'(x) dx$. Passing from the microscopic view to an infinitesimal view, the picture gives two interpretations, leading to the Fundamental Theorem of Calculus:
 
-$$~
+```math
 \int_a^b F'(x) dx = F(b) - F(a).
-~$$
+```
 
 The three theorems of this section, Green's theorem, Stokes' theorem, and the divergence theorem, can all be seen in this manner: the sum of microscopic boundary integrals leads to a macroscopic boundary integral of the entire region; whereas, by reinterpretation, the microscopic boundary integrals are viewed as Riemann sums, which in the limit become integrals of a *related* function over the region.
 
@@ -166,14 +166,14 @@ p
 
 Consider the boundary integral $\oint_c F\cdot\vec{T} ds$ around the smallest (green) squares. We have seen that the *curl* at a point in a direction is given in terms of the limit. Let the plane be the $x-y$ plane, and the $\hat{k}$ direction be the one coming out of the figure. In the derivation of the curl, we saw that the line integral for circulation around the square satisfies:
 
-$$~
+```math
 \lim \frac{1}{\Delta{x}\Delta{y}} \oint_C F \cdot\hat{T}ds =
  \frac{\partial{F_y}}{\partial{x}} - \frac{\partial{F_x}}{\partial{y}}.
-~$$
+```
 
 If the green squares are small enough, then the line integrals satisfy:
 
-$$~
+```math
 \oint_C F \cdot\hat{T}ds
 \approx
 \left(
@@ -181,15 +181,15 @@ $$~
 -
 \frac{\partial{F_x}}{\partial{y}}
 \right) \Delta{x}\Delta{y} .
-~$$
+```
 
 We interpret the right hand side as a Riemann sum approximation for the $2$ dimensional integral of the function $f(x,y) = \frac{\partial{F_x}}{\partial{y}} - \frac{\partial{F_y}}{\partial{x}}=\text{curl}(F)$, the two-dimensional curl. Were the green squares continued to fill out the large blue square, then the sum of these terms would approximate the integral
 
-$$~
+```math
 \iint_S f(x,y) dA = \iint_S
 \left(\frac{\partial{F_y}}{\partial{x}} - \frac{\partial{F_x}}{\partial{y}}\right) dA
 = \iint_S \text{curl}(F) dA.
-~$$
+```
 
 
 However, the microscopic boundary integrals have cancellations that lead to a macroscopic boundary integral.  The sum of $\oint_C F \cdot\hat{T}ds$ over the $4$ green squares will be equal to $\oint_{C_r} F\cdot\hat{T}ds$, where $C_r$ is the red square, as the interior line integral pieces will all cancel off. The sum of $\oint_{C_r} F \cdot\hat{T}ds$ over the $4$ red squares will equal $\oint_{C_b} F \cdot\hat{T}ds$, where $C_b$ is the oriented path around the blue square, as again the interior line pieces will cancel off. Etc.
@@ -217,10 +217,10 @@ Some examples, following Strang, are:
 
 Let $F(x,y) = \langle -y, x\rangle$. Then $\frac{\partial{F_y}}{\partial{x}} - \frac{\partial{F_x}}{\partial{y}}=2$, so
 
-$$~
+```math
 \frac{1}{2}\oint_C F\cdot\hat{T}ds = \frac{1}{2}\oint_C (xdy - ydx) =
 \iint_D dA = A(D).
-~$$
+```
 
 This gives a means to compute the area of a region by integrating around its boundary.
 
@@ -235,29 +235,29 @@ F(v) = F(v...)
 
 r(t) = [a*cos(t),b*sin(t)]
 
-@syms a::positive b::positive
+@syms a::positive b::positive t
 (1//2) * integrate( F(r(t)) ⋅ diff.(r(t),t), (t, 0, 2PI))
 ```
 
 To compute the area of the triangle with vertices $(0,0)$, $(a,0)$ and $(0,b)$ we can orient the boundary counter clockwise. Let $A$ be the line segment from $(0,b)$ to $(0,0)$, $B$ be the line segment from $(0,0)$ to $(a,0)$, and $C$ be the other. Then
 
 
-$$~
+```math
 \begin{align}
 \frac{1}{2} \int_A F\cdot\hat{T} ds &=\frac{1}{2} \int_A -ydx = 0\\
 \frac{1}{2} \int_B F\cdot\hat{T} ds &=\frac{1}{2} \int_B xdy = 0,
 \end{align}
-~$$
+```
 
 as on $A$, $y=0$ and $dy=0$ and on $B$, $x=0$ and $dx=0$.
 
 On $C$ we have $\vec{r}(t) = (0, b) + t\cdot(1,-b/a) =\langle t, b-(bt)/a\rangle$ from $t=a$ to $0$
 
-$$~
+```math
 \int_C F\cdot \frac{d\vec{r}}{dt} dt =
 \int_a^0 \langle -b + (bt)/a), t\rangle\cdot\langle 1, -b/a\rangle dt
 = \int_a^0 -b dt = -bt\mid_{a}^0 = ba.
-~$$
+```
 
 Dividing by $1/2$ give the familiar answer $A=(1/2) a b$.
 
@@ -274,12 +274,17 @@ For the two dimensional case the curl is a scalar. *If* $F = \langle F_x, F_y\ra
 
 Now assume $\partial{F_y}/\partial{x} - \partial{F_x}/\partial{y} = 0$. Let $P$ and $Q$ be two points in the plane. Take any path, $C_1$ from $P$ to $Q$ and any return path, $C_2$, from $Q$ to $P$ that do not cross and such that $C$, the concatenation of the two paths, satisfies Green's theorem. Then, as $F$ is continuous on an open interval containing $D$, we have:
 
-$$~
-0 = \iint_D 0 dA =
-\iint_D \left(\partial{F_y}/\partial{x} - \partial{F_x}/\partial{y}\right)dA =
-\oint_C F \cdot \hat{T} ds =
+```math
+\begin{align*}
+0 &= \iint_D 0 dA \\
+&=
+\iint_D \left(\partial{F_y}/\partial{x} - \partial{F_x}/\partial{y}\right)dA \\
+&=
+\oint_C F \cdot \hat{T} ds \\
+&=
 \int_{C_1} F \cdot \hat{T} ds + \int_{C_2}F \cdot \hat{T} ds.
-~$$
+\end{align*}
+```
 
 Reversing $C_2$ to go from $P$ to $Q$, we see the two work integrals are identical, that is the field is conservative.
 
@@ -293,14 +298,14 @@ For example, let $F(x,y) = \langle \sin(xy), \cos(xy) \rangle$. Is this a conser
 
 We can check by taking partial derivatives. Those of interest are:
 
-$$~
+```math
 \begin{align}
 \frac{\partial{F_y}}{\partial{x}} &= \frac{\partial{(\cos(xy))}}{\partial{x}} =
 -\sin(xy) y,\\
 \frac{\partial{F_x}}{\partial{y}} &= \frac{\partial{(\sin(xy))}}{\partial{y}} =
 \cos(xy)x.
 \end{align}
-~$$
+```
 
 It is not the case that  $\partial{F_y}/\partial{x} - \partial{F_x}/\partial{y}=0$, so this vector field is *not* conservative.
 
@@ -326,9 +331,9 @@ diff(Fy, x) - diff(Fx, y) |> simplify
 
 As the integrand is ``00``,  $\iint_D \left( \partial{F_y}/{\partial{x}}-\partial{F_xy}/{\partial{y}}\right)dA = 0$, as well. But,
 
-$$~
+```math
 F\cdot\hat{T} = \frac{R}{R\cdot{R}} \cdot \frac{R}{R\cdot{R}} = \frac{R\cdot{R}}{(R\cdot{R})^2} = \frac{1}{R\cdot{R}},
-~$$
+```
 
 so $\oint_C F\cdot\hat{T}ds = 2\pi$, $C$ being the unit circle so $R\cdot{R}=1$.
 
@@ -357,24 +362,24 @@ p
 
 Let $A$ label the red line, $B$ the green curve, $C$ the blue line, and $D$ the black line. Then the area is given from Green's theorem by considering half of the the line integral of $F(x,y) = \langle -y, x\rangle$ or $\oint_C (xdy - ydx)$. To that matter we have:
 
-$$~
+```math
 \begin{align}
 \int_A (xdy - ydx) &= a f(a)\\
 \int_C (xdy - ydx) &= b(-f(b))\\
 \int_D (xdy - ydx) &= 0\\
 \end{align}
-~$$
+```
 
 Finally the integral over $B$, using integration by parts:
 
-$$~
+```math
 \begin{align}
 \int_B F(\vec{r}(t))\cdot \frac{d\vec{r}(t)}{dt} dt &=
 \int_b^a \langle -f(t),t)\rangle\cdot\langle 1, f'(t)\rangle dt\\
 &= \int_a^b f(t)dt - \int_a^b tf'(t)dt\\
 &= \int_a^b f(t)dt - \left(tf(t)\mid_a^b - \int_a^b f(t) dt\right).
 \end{align}
-~$$
+```
 
 Combining, we have after cancellation $\oint (xdy - ydx) = 2\int_a^b f(t) dt$, or after dividing by $2$ the signed area under the curve.
 
@@ -403,16 +408,16 @@ The cut leads to a counter-clockwise orientation  on the outer ring and a clockw
 
 To see that the area integral of $F(x,y) = (1/2)\langle -y, x\rangle$ produces the area for this orientation we have, using $C_1$ as the outer ring, and $C_2$ as the inner ring:
 
-$$~
+```math
 \begin{align}
 \oint_{C_1} F \cdot \hat{T} ds &=
-\int_0^{2\pi} (1/2)(2)\langle -\sin(t), \cos(t)\rangle \cdot (2)\langle-\sin(t), \cos(t)\rangle dt
-= (1/2) (2\pi) 4 = 4\pi\\
+\int_0^{2\pi} (1/2)(2)\langle -\sin(t), \cos(t)\rangle \cdot (2)\langle-\sin(t), \cos(t)\rangle dt \\
+&= (1/2) (2\pi) 4 = 4\pi\\
 \oint_{C_2} F \cdot \hat{T} ds &=
 \int_{0}^{2\pi}  (1/2) \langle \sin(t), \cos(t)\rangle \cdot \langle-\sin(t), -\cos(t)\rangle dt\\
 &= -(1/2)(2\pi) = -\pi.
 \end{align}
-~$$
+```
 
 (Using $\vec{r}(t) = 2\langle \cos(t), \sin(t)\rangle$ for the outer ring and $\vec{r}(t) = 1\langle \cos(t), -\sin(t)\rangle$ for the inner ring.)
 
@@ -425,22 +430,22 @@ Green's theorem has a complement in terms of flow across $C$. As $C$ is positive
 
 Let $F = \langle F_x, F_y \rangle$ and $G = \langle F_y, -F_x \rangle$, then $G\cdot\hat{T} = -F\cdot\hat{N}$. The curl formula applied to $G$ becomes
 
-$$~
+```math
 \frac{\partial{G_y}}{\partial{x}} - \frac{\partial{G_x}}{\partial{y}} =
 \frac{\partial{-F_x}}{\partial{x}}-\frac{\partial{(F_y)}}{\partial{y}}
 =
 -\left(\frac{\partial{F_x}}{\partial{x}} + \frac{\partial{F_y}}{\partial{y}}\right)=
 -\nabla\cdot{F}.
-~$$
+```
 
 Green's theorem applied to $G$ then gives this formula for $F$:
 
-$$~
+```math
 \oint_C F\cdot\hat{N} ds =
 -\oint_C G\cdot\hat{T} ds =
 -\iint_D (-\nabla\cdot{F})dA =
 \iint_D \nabla\cdot{F}dA.
-~$$
+```
 
 The right hand side integral is the $2$-dimensional divergence, so this has the interpretation that the flux through $C$ ($\oint_C F\cdot\hat{N} ds$) is the integral of the divergence. (The divergence is defined in terms of a limit of this picture, so this theorem extends the microscopic view to a bigger view.)
 
@@ -522,17 +527,17 @@ Again, the microscopic boundary integrals when added will give a macroscopic bou
 But, as seen in the derivation of the divergence, only modified for $2$ dimensions, we have
 $\nabla\cdot{F} = \lim \frac{1}{\Delta S} \oint_C F\cdot\hat{N}$, so for each cell
 
-$$~
+```math
 \oint_{C_i} F\cdot\hat{N} \approx \left(\nabla\cdot{F}\right)\Delta{x}\Delta{y},
-~$$
+```
 an approximating Riemann sum for $\iint_D \nabla\cdot{F} dA$. This yields:
 
-$$~
+```math
 \oint_C (F \cdot\hat{N}) dA =
 \sum_i \oint_{C_i}  (F \cdot\hat{N}) dA \approx
 \sum \left(\nabla\cdot{F}\right)\Delta{x}\Delta{y} \approx
 \iint_S \nabla\cdot{F}dA,
-~$$
+```
 
 the approximation signs becoming equals signs in the limit.
 
@@ -549,11 +554,11 @@ We have the divergence is simply $a + b$ so $\iint_D (a+b)dA = (a+b)A(D) = 4(a+b
 
 The integral of the flow across $C$ consists of $4$ parts. By symmetry, they all should be similar. We consider the line segment connecting $(1,-1)$ to $(1,1)$ (which has the proper counterclockwise orientation):
 
-$$~
+```math
 \int_C F \cdot \hat{N} ds=
 \int_{-1}^1 \langle F_x, F_y\rangle\cdot\langle 0, 1\rangle ds =
 \int_{-1}^1 b dy = 2b.
-~$$
+```
 
 Integrating across the top will give $2a$, along the bottom $2a$, and along the left side $2b$ totaling $4(a+b)$.
 
@@ -586,10 +591,10 @@ The gradient of $f=\langle v_x, v_y\rangle$ is orthogonal to the contour lines o
 
 As an [example](https://en.wikipedia.org/wiki/Potential_flow#Examples_of_two-dimensional_flows) consider the following in polar coordinates:
 
-$$~
+```math
 f(r, \theta) = A r^n \cos(n\theta),\quad
 g(r, \theta) = A r^n \sin(n\theta).
-~$$
+```
 
 The constant $A$ just sets the scale, the parameter $n$ has a qualitative effect on the contour lines. Consider $n=2$ visualized below:
 
@@ -607,7 +612,7 @@ f(v) = f(v...); g(v)= g(v...)
 xs = ys = range(-2,2, length=50)
 p = contour(xs, ys, f∘Φ, color=:red, legend=false, aspect_ratio=:equal)
 contour!(p, xs, ys, g∘Φ, color=:blue, linewidth=3)
-pyplot()
+#pyplot()
 p
 ```
 
@@ -636,29 +641,34 @@ Imagine if instead of the retro labeling,  a rectangular grid were drawn on the
 
 Now imagine the popcorn expanding, but rather than worry about burning, focusing instead on what happens to the integral of the curl in the direction of the normal, we have
 
-$$~
+```math
 \nabla\times{F} \cdot\hat{N} = \lim \frac{1}{\Delta{S}} \oint_C F\cdot\hat{T} ds
 \approx \frac{1}{\Delta{S}} F\cdot\hat{T} \Delta{s}.
-~$$
+```
 
 This gives the series of approximations:
 
-$$~
-\oint_C F\cdot\hat{T} ds =
-\sum \oint_{C_i} F\cdot\hat{T} ds \approx
-\sum F\cdot\hat{T} \Delta s \approx
-\sum \nabla\times{F}\cdot\hat{N} \Delta{S} \approx
+```math
+\begin{align*}
+\oint_C F\cdot\hat{T} ds &=
+\sum \oint_{C_i} F\cdot\hat{T} ds \\
+&\approx
+\sum F\cdot\hat{T} \Delta s \\
+&\approx
+\sum \nabla\times{F}\cdot\hat{N} \Delta{S} \\
+&\approx
 \iint_S \nabla\times{F}\cdot\hat{N} dS.
-~$$
+\end{align*}
+```
 
 In terms of our expanding popcorn, the boundary integral - after accounting for cancellations, as in Green's theorem - can be seen as a microscopic sum of boundary integrals each of which is approximated by a term
 $\nabla\times{F}\cdot\hat{N} \Delta{S}$ which is viewed as a Riemann sum approximation for the the integral of the curl over the surface. The cancellation depends on a proper choice of orientation, but with that we have:
 
 
 > **Stokes' theorem**: Let $S$ be an orientable smooth surface in $R^3$ with boundary $C$, $C$ oriented so that the chosen normal for $S$ agrees with the right-hand rule for $C$'s orientation. Then *if* $F$ has continuous partial derivatives
-> $$~
+> ```math
 > \oint_C F \cdot\hat{T} ds = \iint_S (\nabla\times{F})\cdot\hat{N} dA.
-> ~$$
+> ```
 
 
 Green's theorem is an immediate consequence upon viewing the region in $R^2$ as a surface in $R^3$ with normal $\hat{k}$.
@@ -707,51 +717,51 @@ integrate(integrandₛ, (t, 0, 2PI))
 
 Ampere's circuital law relates the line integral of the magnetic field to the induced current through:
 
-$$~
+```math
 \oint_C B\cdot\hat{T} ds = \mu_0 I.
-~$$
+```
 
 The goal here is to re-express this integral law to produce a law at each point of the field. Let $S$ be a surface with boundary $C$, Let $J$ be the current density - $J=\rho v$, with $\rho$ the density of the current (not time-varying) and $v$ the velocity. The current can be re-expressed as $I = \iint_S J\cdot\hat{n}dA$. (If the current flows through a wire and $S$ is much bigger than the wire, this is still valid as $\rho=0$ outside of the wire.)
 
 We then have:
 
 
-$$~
+```math
 \mu_0 \iint_S J\cdot\hat{N}dA =
 \mu_0 I =
 \oint_C B\cdot\hat{T} ds =
 \iint_S (\nabla\times{B})\cdot\hat{N}dA.
-~$$
+```
 
 As $S$ and $C$ are arbitrary, this implies the integrands of the surface integrals are equal, or:
 
-$$~
+```math
 \nabla\times{B} = \mu_0 J.
-~$$
+```
 
 
 ##### Example: Faraday's law
 
 (Strang) Suppose $C$ is a wire and there is a time-varying magnetic field $B(t)$. Then Faraday's law says the *flux* passing within $C$ through a surface $S$ with boundary $C$ of the magnetic field, $\phi = \iint B\cdot\hat{N}dS$, induces an electric field $E$ that does work:
 
-$$~
+```math
 \oint_C E\cdot\hat{T}ds = -\frac{\partial{\phi}}{\partial{t}}.
-~$$
+```
 
 Faraday's law is an empirical statement. Stokes' theorem can be used to produce one of Maxwell's equations. For any surface $S$, as above with its boundary being $C$, we have both:
 
-$$~
+```math
 -\iint_S \left(\frac{\partial{B}}{\partial{t}}\cdot\hat{N}\right)dS =
 -\frac{\partial{\phi}}{\partial{t}} =
 \oint_C E\cdot\hat{T}ds =
 \iint_S (\nabla\times{E}) dS.
-~$$
+```
 
 This is true for any capping surface for $C$. Shrinking $C$ to a point means it will hold for each point in $R^3$. That  is:
 
-$$~
+```math
 \nabla\times{E} = -\frac{\partial{B}}{\partial{t}}.
-~$$
+```
 
 
 ##### Example: Conservative fields
@@ -783,20 +793,20 @@ curl(F(x,y,z), [x,y,z])
 
 We need $\phi$ with $\partial{\phi}/\partial{x} = F_x = yz^2$. To that end, we integrate in $x$:
 
-$$~
+```math
 \phi(x,y,z) = \int yz^2 dx = xyz^2 + g(y,z),
-~$$
+```
 the function $g(y,z)$ is a "constant" of integration (it doesn't depend on $x$). That $\partial{\phi}/\partial{x} = F_x$ is true is easy to verify. Now, consider the partial in $y$:
 
-$$~
+```math
 \frac{\partial{\phi}}{\partial{y}} = xz^2 + \frac{\partial{g}}{\partial{y}} = F_y = xz^2.
-~$$
+```
 
 So we have $\frac{\partial{g}}{\partial{y}}=0$ or $g(y,z) = h(z)$, some constant in $y$. Finally, we must have $\partial{\phi}/\partial{z} = F_z$, or
 
-$$~
+```math
 \frac{\partial{\phi}}{\partial{z}} = 2xyz + h'(z) = F_z = 2xyz,
-~$$
+```
 
 So $h'(z) = 0$. This value can be any constant, even $0$ which we take, so that $g(y,z) = 0$ and $\phi(x,y,z) = xyz^2$ is a scalar potential for $F$.
 
@@ -847,18 +857,18 @@ However, the function $F(x,y,z) = \langle x, y,z\rangle/\sqrt{x^2+y^2+z^2}$ has
 The divergence theorem is a consequence of a simple observation. Consider two adjacent cubic regions that share a common face.
 The boundary integral, $\oint_S F\cdot\hat{N} dA$, can be computed for each cube. The surface integral requires a choice of normal, and the convention is to use the outward pointing normal. The common face of the two cubes has *different* outward pointing normals, the difference being a minus sign. As such, the contribution of the surface integral over this face for one cube is *cancelled* out by the contribution of the surface integral over this face for the adjacent cube. As with Green's theorem, this means for a cubic partition, that only the contribution over the boundary is needed to compute the boundary integral. In formulas, if $V$ is a $3$ dimensional cubic region with boundary $S$ and it is partitioned into smaller cubic subregions, $V_i$ with surfaces $S_i$, we have:
 
-$$~
+```math
 \oint_S F\cdot{N} dA = \sum \oint_{S_i} F\cdot{N} dA.
-~$$
+```
 
 If the partition provides a microscopic perspective, then the divergence approximation $\nabla\cdot{F} \approx (1/\Delta{V_i}) \oint_{S_i} F\cdot{N} dA$ can be used to say:
 
-$$~
+```math
 \oint_S F\cdot{N} dA =
 \sum \oint_{S_i} F\cdot{N} dA \approx
 \sum (\nabla\cdot{F})\Delta{V_i} \approx
 \iiint_V \nabla\cdot{F} dV,
-~$$
+```
 
 the last approximation through a Riemann sum approximation. This heuristic leads to:
 
@@ -965,29 +975,29 @@ The divergence theorem provides two means to compute a value, the point here is
 
 Following Schey, we now consider a continuous analog to the crowd counting problem through a flow with a non-uniform density that may vary in time. Let $\rho(x,y,z;t)$ be the time-varying density and $v(x,y,z;t)$ be a vector field indicating the direction of flow. Consider some three-dimensional volume, $V$, with boundary $S$ (though two-dimensional would also be applicable). Then these integrals have interpretations:
 
-$$~
+```math
 \begin{align}
 \iiint_V \rho dV &&\quad\text{Amount contained within }V\\
 \frac{\partial}{\partial{t}} \iiint_V \rho dV &=
 \iiint_V \frac{\partial{\rho}}{\partial{t}} dV &\quad\text{Change in time of amount contained within }V
 \end{align}
-~$$
+```
 
 Moving the derivative inside the integral requires an assumption of continuity.
 Assume the material is *conserved*, meaning that if the amount in the volume $V$ changes it must  flow in and out through the boundary. The flow out through $S$, the boundary of $V$, is
 
-$$~
+```math
 \oint_S (\rho v)\cdot\hat{N} dS,
-~$$
+```
 
 using the customary outward pointing normal for the orientation of $S$.
 
 So we have:
 
-$$~
+```math
 \iiint_V \frac{\partial{\rho}}{\partial{t}} dV =
 -\oint_S (\rho v)\cdot\hat{N} dS = - \iiint_V \nabla\cdot\left(\rho v\right)dV.
-~$$
+```
 
 The last equality by the divergence theorem, the minus sign as a positive change in amount within $V$ means flow *opposite* the outward pointing normal for $S$.
 
@@ -995,9 +1005,9 @@ The volume $V$ was arbitrary. While it isn't the case that two integrals being e
 
 That is, under the *assumptions* that material is conserved and density is continuous a continuity equation can be derived from the divergence theorem:
 
-$$~
+```math
 \nabla\cdot(\rho v) = - \frac{\partial{\rho}}{dt}.
-~$$
+```
 
 
 
@@ -1018,11 +1028,11 @@ The simplification done by SymPy masks the presence of $R^{-5/2}$ when taking th
 
 *Were* the divergence theorem applicable, then the integral of $F$ over the unit sphere would mean:
 
-$$~
+```math
 0 = \iiint_V \nabla\cdot{F} dV =
 \oint_S F\cdot{N}dS = \oint_S \frac{R}{\|R\|^3} \cdot{R} dS =
 \oint_S 1 dS = 4\pi.
-~$$
+```
 
 Clearly, as $0$ is not equal to $4\pi$, the divergence theorem can not apply.
 
@@ -1045,8 +1055,8 @@ L"We must have $\text{curl}(F) = 1$",
 L"We must have $\text{curl}(F) = 0$",
 L"We must have $\text{curl}(F) = x$"
 ]
-ans = 1
-radioq(choices, ans)
+answ = 1
+radioq(choices, answ)
 ```
 
 
@@ -1074,8 +1084,8 @@ raw" ``(1/2) \int r d\theta``",
 raw" ``\int r^2 d\theta``",
 raw" ``(1/2) \int r^2d\theta``"
 ]
-ans=4
-radioq(choices, ans)
+answ=4
+radioq(choices, answ)
 ```
 
 
@@ -1094,8 +1104,8 @@ raw" ``3\pi/8``",
 raw" ``\pi/4``",
 raw" ``\pi/2``"
 ]
-ans = 1
-radioq(choices, ans)
+answ = 1
+radioq(choices, answ)
 ```
 
 ###### Question
@@ -1110,8 +1120,8 @@ raw" ``0``",
 raw" ``1``",
 raw" ``2``"
 ]
-ans =1
-radioq(choices, ans, keep_order=true)
+answ =1
+radioq(choices, answ, keep_order=true)
 ```
 
 As the curl is a constant, say $c$, we have $\iint_S (\nabla\times{F}) dS = c \cdot 1$. This is?
@@ -1122,8 +1132,8 @@ raw" ``0``",
 raw" ``1``",
 raw" ``2``"
 ]
-ans =1
-radioq(choices, ans, keep_order=true)
+answ =1
+radioq(choices, answ, keep_order=true)
 ```
 
 To integrate around the boundary we have ``4`` terms: the path $A$ connecting $(0,0)$ to $(1,0)$ (on the $x$ axis), the path $B$ connecting $(1,0)$ to $(1,1)$, the path $C$ connecting $(1,1)$ to $(0,1)$, and the path $D$ connecting $(0,1)$ to $(0,0)$ (along the $y$ axis).
@@ -1132,8 +1142,8 @@ Which path has tangent $\hat{j}$?
 
 ```julia; hold=true; echo=false
 choices = ["`` A``","`` B``"," ``C``"," ``D``"]
-ans = 2
-radioq(choices, ans, keep_order=true)
+answ = 2
+radioq(choices, answ, keep_order=true)
 ```
 
 Along  path $C$, $F(x,y) = [1,x]$ and $\hat{T}=-\hat{i}$ so $F\cdot\hat{T} = -1$. The path integral $\int_C (F\cdot\hat{T})ds = -1$. What is the value of the path integral over $A$?
@@ -1144,8 +1154,8 @@ raw" ``-1``",
 raw" ``0``",
 raw" ``1``"
 ]
-ans = 3
-radioq(choices, ans, keep_order=true)
+answ = 3
+radioq(choices, answ, keep_order=true)
 ```
 
 What is the integral over the oriented boundary of $S$?
@@ -1156,8 +1166,8 @@ raw" ``0``",
 raw" ``1``",
 raw" ``2``"
 ]
-ans =1
-radioq(choices, ans, keep_order=true)
+answ =1
+radioq(choices, answ, keep_order=true)
 ```
 
 
@@ -1171,8 +1181,8 @@ choices = [
 L"They are the same, as Green's theorem applies to the area, $S$, between $C_1$ and $C_2$ so $\iint_S \nabla\cdot{F}dA = 0$."
 L"They  differ by a minus sign, as Green's theorem applies to the area, $S$, between $C_1$ and $C_2$ so $\iint_S \nabla\cdot{F}dA = 0$."
 ]
-ans = 1
-radioq(choices, ans)
+answ = 1
+radioq(choices, answ)
 ```
 
 ###### Question
@@ -1185,8 +1195,8 @@ choices = [
 L"Also $2\pi$, as Green's theorem applies to the region formed by the square minus the circle and so the overall flow integral around the boundary is $0$, so the two will be the same.",
 L"It is $-2\pi$, as Green's theorem applies to the region formed by the square minus the circle and so the overall flow integral around the boundary is $0$, so the two will have opposite signs, but the same magnitude."
 ]
-ans = 1
-radioq(choices, ans)
+answ = 1
+radioq(choices, answ)
 ```
 
 ###### Question
@@ -1199,8 +1209,8 @@ raw" ``4/3 \pi``",
 raw" ``4\pi``",
 raw" ``\pi``"
 ]
-ans = 1
-radioq(choices, ans)
+answ = 1
+radioq(choices, answ)
 ```
 
 
@@ -1214,8 +1224,8 @@ raw" ``1``",
 raw" ``2``",
 raw" ``3``"
 ]
-ans = 1
-radioq(choices, ans, keep_order=true)
+answ = 1
+radioq(choices, answ, keep_order=true)
 ```
 
 
@@ -1229,8 +1239,8 @@ raw" ``\log(\rho)``",
 raw" ``1/\rho``",
 raw" ``\rho``"
 ]
-ans = 1
-radioq(choices, ans)
+answ = 1
+radioq(choices, answ)
 ```
 
 Based on this information, for $S$ a surface not including the origin with boundary $C$, a simple closed curve, what is $\oint_C F\cdot\hat{T}ds$?
@@ -1240,8 +1250,8 @@ choices = [
 L"It is $0$, as, by Stoke's theorem, it is equivalent to $\iint_S (\nabla\times\nabla{\phi})dS = \iint_S 0 dS = 0$.",
 L"It is $2\pi$, as this is the circumference of the unit circle"
 ]
-ans = 1
-radioq(choices, ans)
+answ = 1
+radioq(choices, answ)
 ```
 
 ###### Question
@@ -1254,8 +1264,8 @@ raw" ``2\pi``",
 raw" ``2``",
 raw" ``0``"
 ]
-ans = 1
-radioq(choices, ans)
+answ = 1
+radioq(choices, answ)
 ```
 
 ###### Question
@@ -1274,8 +1284,8 @@ choices = [
 "the field is *not* conservative.",
 "the field *is* conservative"
 ]
-ans=1
-radioq(choices, ans)
+answ=1
+radioq(choices, answ)
 ```
 
 ###### Question
@@ -1311,8 +1321,8 @@ choices = [
 "The maximum number in a row",
 "The row number plus 1"
 ]
-ans = 1
-radioq(choices, ans)
+answ = 1
+radioq(choices, answ)
 ```
 
 ###### Question
@@ -1321,10 +1331,10 @@ radioq(choices, ans)
 
 In 1846, Cauchy proved
 
-$$~
+```math
 \int\left(p\frac{dx}{ds} + q \frac{dy}{ds}\right)ds =
 \pm\iint\left(\frac{\partial{p}}{\partial{y}} - \frac{\partial{q}}{\partial{x}}\right)dx dy.
-~$$
+```
 
 This is a form of:
 
@@ -1334,6 +1344,6 @@ choices = [
 "The divergence (Gauss') theorem",
 "Stokes' theorem"
 ]
-ans = 1
-radioq(choices, ans, keep_order=true)
+answ = 1
+radioq(choices, answ, keep_order=true)
 ```